I bought a 1 Ohm 10watt ceramic resistor from Radio Shack. I'm wondering how
much surge current these can handle. For continuos duty, it 10 amps.
temperature coefficient is 0 +/- 800ppm (typ.) I know what TC is, but what's
the 800ppm mean here?
Thermal strength is 5lbs minimum. What is this?
Max working voltage is Sq, root of PR. Is that the square root of power time
resistance? That's only 3.16. Can't be right.
Actually, 10 amps through 1 ohm is 100 watts. If it is rated for 10 watts,
then since power = I^2*R and R=1, then 10 watts = I^2. Therefore I=sqrt(10)
For 'surge', it would depend on a lot of things, like the total mass of the
resistor and its heat-capacity, and just what you mean by 'surge'. Short
pulse that doesn't exceed the ampacity of the leads and doesn't overheat the
central core, and is infrequent enough to allow cooling down between surges
could be tolerated.
Parts Per Million. Its resistance will change 800 parts per million. In
this case, with a 1 ohm resistor, that would be 800/1 000 000 ohm. If it
were a 20 ohm resistor, it would change 20* 800/1 000 000.
Well, since power = E^2/R, and R=1,
then 10 watts = E^2/1 or E = sqrt(10/1) = ~3.16
Yes, that's right. Of course, some engineers will tell you, you shouldn't
expect to operate such a resistor continuously right at its power rating for
a long time. Better to get a higher wattage resistor and operate less than
The thing to remember is that power rating has everything to do with how the
resistor gets rid of its heat. A rule of thumb is that you'll get something
like 150 degrees C temp rise for each watt you force through a square inch for
'normal' convective heat transfer. If there's forced air cooling your resistor
can probably handle more than a watt before its core fails, because forcing air
movement changes that 150 degree number down by a significant amount.
I think the number you are using is a bit high.
Westing house built a transformer 3 inches on a side which would be 54 inches
of surface area.
At 25,000 watts and 1% loss that would be about 250 watts in the transformer
and as I remember it the surface rise would have been to about 100 degrees C.
Of course that was not acceptable as the internal temp would be much higher.
I had a magic number like yours but can'r remember it.
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i think you'll find that the max continuous current will be 3.16 amps.
at 3.16 amps you'll have 3.16 volts across it.
3.16 amps times 3.16 volts equals about 10 watts.
as to the max current in a surge, it depends greatly on the length of the
pulse and the temperature of the resistor.
you can go higher in current and shorter in time 'til the wire itself
the 800 ppm means 800 parts per million or .0008 ohms for a one ohm
I thought about your surge current question a bit. If it were my analysis to
do, I'd approach it like this. I'd determine the volume of the resistor and
guess at its heat capacity -- that is, how many degrees temp rise you get for a
given impulse of energy. I'd assume you'd not want to see the resistor get
hotter than say 70 degrees C -- maybe as much as 100 degrees is OK, but not
much more than that. So, so long as the average power is 10 watts, I'd guess
you could hit it with any current up to lead melting levels, or voltages up to
arc over or breakdown, so long as the duty cycle is such that the average power
is 10 watts. A hundred amps is 10KW steady state, but if the the on time is a
microsecond and the off time 1 mS, the average power will be 10 watts, the
voltages will be reasonable, and the 'impulse" of energy should not overly
spike the temp of the resistor. On the other hand if the 100 amps were imposed
for a second, you wouldn't have to think about a 1000 second off time, the
resistor would be more than well done.